银行排队办业务问题

使用了 队列和结构体

算是比较麻烦的题了吧（可能我比较菜）

Suppose a bank has *N* windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. The rules for the customers to wait in line are:

- The space inside the yellow line in front of each window is enough to contain a line with
*M*customers. Hence when all the*N*lines are full, all the customers after (and including) the (*NM*+1)st one will have to wait in a line behind the yellow line. - Each customer will choose the shortest line to wait in when crossing the yellow line. If there are two or more lines with the same length, the customer will always choose the window with the smallest number.
*Customeri*will take*Ti*minutes to have his/her transaction processed.- The first
*N*customers are assumed to be served at 8:00am.

Now given the processing time of each customer, you are supposed to tell the exact time at which a customer has his/her business done.

For example, suppose that a bank has 2 windows and each window may have 2 customers waiting inside the yellow line. There are 5 customers waiting with transactions taking 1, 2, 6, 4 and 3 minutes, respectively. At 08:00 in the morning, *customer1* is served at *window1* while *customer2* is served at *window2*. *customer3* will wait in front of *window1* and *customer4* will wait in front of *window2*. *customer5* will wait behind the yellow line.

At 08:01, *customer1* is done and *customer5* enters the line in front of *window1* since that line seems shorter now. *Customer2* will leave at 08:02, *customer4* at 08:06, *customer3* at 08:07, and finally *customer5* at 08:10.

## Input Specification:

Each input file contains one test case. Each case starts with a line containing 4 positive integers: *N* (≤20, number of windows), *M* (≤10, the maximum capacity of each line inside the yellow line), *K* (≤1000, number of customers), and *Q* (≤1000, number of customer queries).

The next line contains *K* positive integers, which are the processing time of the *K* customers.

The last line contains *Q* positive integers, which represent the customers who are asking about the time they can have their transactions done. The customers are numbered from 1 to *K*.

## Output Specification:

For each of the *Q* customers, print in one line the time at which his/her transaction is finished, in the format `HH:MM`

where `HH`

is in [08, 17] and `MM`

is in [00, 59]. Note that since the bank is closed everyday after 17:00, for those customers who cannot be served before 17:00, you must output `Sorry`

instead.

## Sample Input:

```
2 2 7 5
1 2 6 4 3 534 2
3 4 5 6 7
```

## Sample Output:

```
08:07
08:06
08:10
17:00
Sorry
```

## 实现思想

银行排队问题

有N个窗口，每个窗口可排M个人

第N*M+1个及之后的要在黄线后等候，哪个窗口有人走并且人最少，去哪个窗口排队

四个窗口成为四个结构体对象，并存放在`window`

中

`poptime`

为每个窗口第一个人离开的时间——用来判断应该去那个窗口排队

`endtime`

为每个窗口最后一个人结束的时间——用来判断后面的人是否在17：00后输出`Sorry`

先将N*M个人的时间记录，然后再输入后面等待排队的人，并判断是`Sorry`

是否为`true`

## 代码实现

```
#include <cstdio>
#include <queue>
#include <vector>
using namespace std;
struct node {
int poptime, endtime;
queue<int> q;
};
int main() {
int n, m, k, q, index = 1;
scanf("%d%d%d%d", &n, &m, &k, &q);
vector<int> time(k + 1), result(k + 1);
for(int i = 1; i <= k; i++) {
scanf("%d", &time[i]);
}
vector<node> window(n + 1);
vector<bool> sorry(k + 1, false);
for(int i = 1; i <= m; i++) {
for(int j = 1; j <= n; j++) {
if(index <= k) {
window[j].q.push(time[index]);
if(window[j].endtime >= 540){
sorry[index] = true;
}
window[j].endtime += time[index];
if(i == 1)
window[j].poptime = window[j].endtime;
result[index] = window[j].endtime;
index++;
}
}
}
while(index <= k) {
int tempmin = window[1].poptime, tempwindow =1;
for(int i = 2; i <= n; i++) {
if(window[i].poptime < tempmin) {
tempwindow = i;
tempmin = window[i].poptime;
}
}
window[tempwindow].q.pop();
window[tempwindow].q.push(time[index]);
window[tempwindow].poptime += window[tempwindow].q.front();
if(window[tempwindow].endtime >= 540)
sorry[index] = true;
window[tempwindow].endtime += time[index];
result[index] = window[tempwindow].endtime;
index++;
}
for(int i = 1; i <= q; i++) {
int query, minute;
scanf("%d", &query);
minute = result[query];
if(sorry[query] == true) {
printf("Sorry\n");
} else {
printf("%02d:%02d\n", (minute + 480) / 60, (minute + 480) % 60);
}
}
return 0;
}
```

**本文链接：**https://blog.zjgsujz.cn/2020/07/22/pat-advanced-1014/**版权声明：**本博客所有文章除特别声明外，均默认采用 许可协议。